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And, well let's see.
So, before we actually start

8
00:00:25 --> 00:00:30
reviewing for the test,
I still have to tell you a few

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00:00:30 --> 00:00:34
small things because I promised
to say a few words about what's

10
00:00:34 --> 00:00:37
the difference,
or precisely, 

11
00:00:37 --> 00:00:41
what's the difference between
curl being zero and a field

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00:00:41 --> 00:00:45
being a gradient field,
and why we have this assumption

13
00:00:45 --> 00:00:49
that our vector field had to be
defined everywhere for a field

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00:00:49 --> 00:00:53
with curl zero to actually be
conservative for our test for

15
00:00:53 --> 00:01:04
gradient fields to be valid?
So -- More about validity of

16
00:01:04 --> 00:01:17
Green's theorem and things like
that.

17
00:01:17 --> 00:01:29
So, we've seen the statement of
Green's theorem in two forms.

18
00:01:29 --> 00:01:33
Both of them have to do with
comparing a line integral along

19
00:01:33 --> 00:01:38
a closed curve to a double
integral over the region inside

20
00:01:38 --> 00:01:41
enclosed by the curve.
So, 

21
00:01:41 --> 00:01:45
one of them says the line
integral for the work done by a

22
00:01:45 --> 00:01:50
vector field along a closed
curve counterclockwise is equal

23
00:01:50 --> 00:01:54
to the double integral of a curl
of a field over the enclosed

24
00:01:54 --> 00:01:58
region.
And, the other one says the

25
00:01:58 --> 00:02:05
total flux out of the region,
so, the flux through the curve

26
00:02:05 --> 00:02:11
is equal to the double integral
of divergence of a field in the

27
00:02:11 --> 00:02:13
region.
So, in both cases, 

28
00:02:13 --> 00:02:18
we need the vector field to be
defined not only,

29
00:02:18 --> 00:02:21
I mean, the left hand side
makes sense if a vector field is

30
00:02:21 --> 00:02:24
just defined on the curve
because it's just a line

31
00:02:24 --> 00:02:27
integral on C.
We don't care what happens

32
00:02:27 --> 00:02:29
inside.
But, for the right-hand side to

33
00:02:29 --> 00:02:31
make sense,
and therefore for the equality

34
00:02:31 --> 00:02:34
to make sense,
we need the vector field to be

35
00:02:34 --> 00:02:38
defined everywhere inside the
region.

36
00:02:38 --> 00:02:41
So, I said, if there is a point
somewhere in here where my

37
00:02:41 --> 00:02:45
vector field is not defined,
then it doesn't work.

38
00:02:45 --> 00:02:53
And actually,
we've seen that example.

39
00:02:53 --> 00:03:08
So, this only works if F and
its derivatives are defined

40
00:03:08 --> 00:03:17
everywhere in the region,
R.

41
00:03:17 --> 00:03:20
Otherwise, we are in trouble.
OK, 

42
00:03:20 --> 00:03:29
so we've seen for example that
if I gave you the vector field

43
00:03:29 --> 00:03:35
minus yi xj over x squared plus
y squared,

44
00:03:35 --> 00:03:40
so that's the same vector field
that was on that problem set a

45
00:03:40 --> 00:03:45
couple of weeks ago.
Then, well, f is not defined at

46
00:03:45 --> 00:03:52
the origin, but it's defined
everywhere else.

47
00:03:52 --> 00:04:03
And, wherever it's defined,
it's curl is zero.

48
00:04:03 --> 00:04:14
I should say everywhere it's --
And so, if we have a closed

49
00:04:14 --> 00:04:24
curve in the plane,
well, there's two situations.

50
00:04:24 --> 00:04:28
One is if it does not enclose
the origin.

51
00:04:28 --> 00:04:30
Then, 
yes, 

52
00:04:30 --> 00:04:36
we can apply Green's theorem
and it will tell us that it's

53
00:04:36 --> 00:04:42
equal to the double integral in
here of curl F dA,

54
00:04:42 --> 00:04:46
which will be zero because this
is zero.

55
00:04:46 --> 00:04:51
However, 
if I have a curve that encloses

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00:04:51 --> 00:04:54
the origin,
let's say like this, 

57
00:04:54 --> 00:05:03
for example, 
then, 

58
00:05:03 --> 00:05:07
well, I cannot use the same
method because the vector field

59
00:05:07 --> 00:05:11
and its curl are not defined at
the origin.

60
00:05:11 --> 00:05:14
And, in fact, 
you know that ignoring the

61
00:05:14 --> 00:05:17
problem and saying,
well, the curl is still zero

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00:05:17 --> 00:05:19
everywhere,
will give you the wrong answer

63
00:05:19 --> 00:05:23
because we've seen an example.
We've seen that along the unit

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00:05:23 --> 00:05:27
circle the total work is 2 pi
not zero.

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00:05:27 --> 00:05:35
So, we can't use Green.
However, we can't use it

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00:05:35 --> 00:05:38
directly.
So, there is an extended

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00:05:38 --> 00:05:44
version of Green's theorem that
tells you the following thing.

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00:05:44 --> 00:05:49
Well, 
it tells me that even though I

69
00:05:49 --> 00:05:54
can't do things for just this
region enclosed by C prime,

70
00:05:54 --> 00:05:58
I can still do things for the
region in between two different

71
00:05:58 --> 00:06:06
curves.
OK, so let me show you what I

72
00:06:06 --> 00:06:11
have in mind.
So, let's say that I have my

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00:06:11 --> 00:06:15
curve C'.
Where's my yellow chalk?

74
00:06:15 --> 00:06:22
Oh, here.
So, I have this curve C'.

75
00:06:22 --> 00:06:30
I can't apply Green's theorem
inside it, but let's get out the

76
00:06:30 --> 00:06:34
smaller thing.
So, that one I'm going to make

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00:06:34 --> 00:06:41
going clockwise.
You will see why.

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00:06:41 --> 00:06:48
Then, I could say,
well, let me change my mind.

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00:06:48 --> 00:06:49
This picture is not very well
prepared.

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00:06:49 --> 00:06:53
That's because my writer is on
strike.

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00:06:53 --> 00:07:02
OK, so let's say we have C' and
C'' both going counterclockwise.

82
00:07:02 --> 00:07:05
Then, 
I claim that Green's theorem

83
00:07:05 --> 00:07:09
still applies,
and tells me that the line

84
00:07:09 --> 00:07:14
integral along C prime minus the
line integral along C double

85
00:07:14 --> 00:07:19
prime is equal to the double
integral over the region in

86
00:07:19 --> 00:07:23
between.
So here, now,

87
00:07:23 --> 00:07:35
it's this region with the hole
of the curve.

88
00:07:35 --> 00:07:41
And, well, in our case,
that will turn out to be zero

89
00:07:41 --> 00:07:45
because curl is zero.
OK, so this doesn't tell us

90
00:07:45 --> 00:07:47
what each of these two line
integrals is.

91
00:07:47 --> 00:07:49
But actually,
it tells us that they are equal

92
00:07:49 --> 00:07:51
to each other.
And so, by computing one, 

93
00:07:51 --> 00:07:53
you can see actually that for
this vector field,

94
00:07:53 --> 00:07:57
if you take any curve that goes
counterclockwise around the

95
00:07:57 --> 00:08:00
origin,
you would get two pi no matter

96
00:08:00 --> 00:08:04
what the curve is.
So how do you get to this?

97
00:08:04 --> 00:08:07
Why is this not like
conceptually a new theorem?

98
00:08:07 --> 00:08:13
Well, just think of the
following thing.

99
00:08:13 --> 00:08:17
I'm not going to do it on top
of that because it's going to be

100
00:08:17 --> 00:08:23
messy if I draw too many things.
But, so here I have my C''.

101
00:08:23 --> 00:08:29
Here, I have C'.
Let me actually make a slit

102
00:08:29 --> 00:08:33
that will connect them to each
other like this.

103
00:08:33 --> 00:08:37
So now if I take, 
see, 

104
00:08:37 --> 00:08:43
I can form a single closed
curve that will enclose all of

105
00:08:43 --> 00:08:49
this region with kind of an
infinitely thin slit here

106
00:08:49 --> 00:08:52
counterclockwise.
And so, if I go

107
00:08:52 --> 00:08:55
counterclockwise around this
region, basically I go

108
00:08:55 --> 00:08:58
counterclockwise along the outer
curve.

109
00:08:58 --> 00:09:01
Then I go along the slit.
Then I go clockwise along the

110
00:09:01 --> 00:09:04
inside curve,
then back along the slit.

111
00:09:04 --> 00:09:07
And then I'm done.
So, 

112
00:09:07 --> 00:09:11
if I take the line integral
along this big curve consisting

113
00:09:11 --> 00:09:15
of all these pieces,
now I can apply Green's theorem

114
00:09:15 --> 00:09:19
to that because it is the usual
counterclockwise curve that goes

115
00:09:19 --> 00:09:22
around a region where my field
is well-defined.

116
00:09:22 --> 00:09:27
See, I've eliminated the origin
from the picture.

117
00:09:27 --> 00:09:37
And, so the total line integral
for this thing is equal to the

118
00:09:37 --> 00:09:46
integral along C prime,
I guess the outer one.

119
00:09:46 --> 00:09:50
Then, I also need to have what
I do along the inner side.

120
00:09:50 --> 00:09:52
And, the inner side is going to
be C double prime,

121
00:09:52 --> 00:09:57
but going backwards because now
I'm going clockwise on C prime

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00:09:57 --> 00:10:01
so that I'm going
counterclockwise around the

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00:10:01 --> 00:10:04
shaded region.
Well, of course there will be

124
00:10:04 --> 00:10:06
contributions from the line
integral along this wide

125
00:10:06 --> 00:10:08
segment.
But, I do it twice,

126
00:10:08 --> 00:10:17
once each way.
So, they cancel out.

127
00:10:17 --> 00:10:21
So, the white segments cancel
out.

128
00:10:21 --> 00:10:23
You probably shouldn't,
in your notes,

129
00:10:23 --> 00:10:25
write down white segments
because probably they are not

130
00:10:25 --> 00:10:29
white on your paper.
But, hopefully you get the

131
00:10:29 --> 00:10:33
meaning of what I'm trying to
say.

132
00:10:33 --> 00:10:36
OK, so basically that tells
you, you can still play tricks

133
00:10:36 --> 00:10:39
with Green's theorem when the
region has holes in it.

134
00:10:39 --> 00:10:44
You just had to be careful and
somehow subtract some other

135
00:10:44 --> 00:10:48
curve so that together things
will work out.

136
00:10:48 --> 00:10:51
There is a similar thing with
the divergence theorem,

137
00:10:51 --> 00:10:55
of course, with flux and double
integral of div f,

138
00:10:55 --> 00:10:58
you can apply exactly the same
argument.

139
00:10:58 --> 00:11:02
OK, so basically you can apply
Green's theorem for a region

140
00:11:02 --> 00:11:04
that has several boundary
curves.

141
00:11:04 --> 00:11:07
You just have to be careful
that the outer boundary must go

142
00:11:07 --> 00:11:13
counterclockwise.
The inner boundary either goes

143
00:11:13 --> 00:11:19
clockwise, or you put a minus
sign.

144
00:11:19 --> 00:11:26
OK, 
and the last cultural note, 

145
00:11:26 --> 00:11:34
so, the definition, 
we say that a region in the

146
00:11:34 --> 00:11:36
plane,
sorry, I should say a connected

147
00:11:36 --> 00:11:45
region in the plane,
so that means -- So, 

148
00:11:45 --> 00:11:47
connected means it consists of
a single piece.

149
00:11:47 --> 00:11:50
OK, so, connected,
there is a single piece.

150
00:11:50 --> 00:11:53
These two guys together are not
connected.

151
00:11:53 --> 00:11:58
But, if I join them,
then this is a connected

152
00:11:58 --> 00:12:08
region.
We say it's simply connected --

153
00:12:08 --> 00:12:17
-- if any closed curve in it,
OK, 

154
00:12:17 --> 00:12:18
so I need to gave a name to my
region,

155
00:12:18 --> 00:12:22
let's say R, 
any closed curve in R, 

156
00:12:22 --> 00:12:29
bounds, 
no, 

157
00:12:29 --> 00:12:37
sorry.
If the interior of any closed

158
00:12:37 --> 00:12:49
curve in R -- -- is also
contained in R.

159
00:12:49 --> 00:12:51
So, concretely,
what does that mean?

160
00:12:51 --> 00:12:57
That means the region,
R, does not have any holes

161
00:12:57 --> 00:13:02
inside it.
Maybe I should draw two

162
00:13:02 --> 00:13:08
pictures to explain what I mean.
So, 

163
00:13:08 --> 00:13:17
this guy here is simply
connected while -- -- this guy

164
00:13:17 --> 00:13:29
here is not simply connected
because if I take this curve,

165
00:13:29 --> 00:13:34
that's a curve inside my region.
But, the piece that it bounds

166
00:13:34 --> 00:13:38
is not actually entirely
contained in my origin.

167
00:13:38 --> 00:13:41
And, so why is that relevant?
Well, 

168
00:13:41 --> 00:13:45
if you know that your vector
field is defined everywhere in a

169
00:13:45 --> 00:13:47
simply connected region,
then you don't have to worry

170
00:13:47 --> 00:13:50
about this question of,
can I apply Green's theorem to

171
00:13:50 --> 00:13:52
the inside?
You know it's automatically OK

172
00:13:52 --> 00:13:54
because if you have a closed
curve,

173
00:13:54 --> 00:13:59
then the vector field is, 
I mean, if a vector field is

174
00:13:59 --> 00:14:03
defined on the curve it will
also be defined inside.

175
00:14:03 --> 00:14:11
OK, 
so if the domain of definition

176
00:14:11 --> 00:14:25
-- -- of a vector field is
defined and differentiable -- --

177
00:14:25 --> 00:14:38
is simply connected -- -- then
we can always apply -- --

178
00:14:38 --> 00:14:47
Green's theorem -- -- and,
of course, 

179
00:14:47 --> 00:14:49
provided that we do it on a
curve where the vector field is

180
00:14:49 --> 00:14:50
defined.
I mean, your line integral

181
00:14:50 --> 00:14:53
doesn't make sense so there's
nothing to compute.

182
00:14:53 --> 00:14:56
But, if you have,
so, again, the argument would

183
00:14:56 --> 00:14:59
be, well, if a vector field is
defined on the curve,

184
00:14:59 --> 00:15:01
it's also defined inside.
So, 

185
00:15:01 --> 00:15:04
see, 
the problem with that vector

186
00:15:04 --> 00:15:07
field here is precisely that its
domain of definition is not

187
00:15:07 --> 00:15:09
simply connected because there
is a hole,

188
00:15:09 --> 00:15:17
namely the origin.
OK, so for this guy,

189
00:15:17 --> 00:15:28
domain of definition,
which is plane minus the origin

190
00:15:28 --> 00:15:39
with the origin removed is not
simply connected.

191
00:15:39 --> 00:15:42
And so that's why you have this
line integral that makes perfect

192
00:15:42 --> 00:15:45
sense, but you can't apply
Green's theorem to it.

193
00:15:45 --> 00:15:47
So now, what does that mean a
particular?

194
00:15:47 --> 00:15:51
Well, we've seen this criterion
that if a curl of the vector

195
00:15:51 --> 00:15:55
field is zero and it's defined
in the entire plane,

196
00:15:55 --> 00:15:58
then the vector field is
conservative,

197
00:15:58 --> 00:16:01
and it's a gradient field.
And, the argument to prove that

198
00:16:01 --> 00:16:03
is basically to use Green's
theorem.

199
00:16:03 --> 00:16:07
So, in fact, 
the actual optimal statement

200
00:16:07 --> 00:16:11
you can make is if a vector
field is defined in a simply

201
00:16:11 --> 00:16:13
connected region,
and its curl is zero, 

202
00:16:13 --> 00:16:26
then it's a gradient field.
So, let me just write that down.

203
00:16:26 --> 00:16:29
So, the correct statement,
I mean, the previous one we've

204
00:16:29 --> 00:16:35
seen is also correct.
But this one is somehow better

205
00:16:35 --> 00:16:45
and closer to what exactly is
needed if curl F is zero and the

206
00:16:45 --> 00:16:55
domain of definition where F is
defined is simply connected --

207
00:16:55 --> 00:17:04
-- then F is conservative.
And that means also it's a

208
00:17:04 --> 00:17:11
gradient field.
It's the same thing.

209
00:17:11 --> 00:17:23
OK, any questions on this?
No?

210
00:17:23 --> 00:17:27
OK, some good news.
What I've just said here won't

211
00:17:27 --> 00:17:31
come up on the test on Thursday.
OK.

212
00:17:31 --> 00:17:35
(APPLAUSE) Still,
it's stuff that you should be

213
00:17:35 --> 00:17:39
aware of generally speaking
because it will be useful,

214
00:17:39 --> 00:17:42
say, on the next week's problem
set.

215
00:17:42 --> 00:17:46
And, 
maybe on the final it would be, 

216
00:17:46 --> 00:17:48
there won't be any really, 
really complicated things

217
00:17:48 --> 00:17:53
probably,
but you might need to be at

218
00:17:53 --> 00:18:01
least vaguely aware of this
issue of things being simply

219
00:18:01 --> 00:18:04
connected.
And by the way,

220
00:18:04 --> 00:18:08
I mean, this is also somehow
the starting point of topology,

221
00:18:08 --> 00:18:12
which is the branch of math
that studies the shapes of

222
00:18:12 --> 00:18:13
regions.
So, 

223
00:18:13 --> 00:18:15
in particular,
you can try to distinguish

224
00:18:15 --> 00:18:18
domains in the plains by looking
at whether they're simply

225
00:18:18 --> 00:18:21
connected or not,
and what kinds of features they

226
00:18:21 --> 00:18:25
have in terms of how you can
joint point what kinds of curves

227
00:18:25 --> 00:18:28
exist in them.
And, since that's the branch of

228
00:18:28 --> 00:18:32
math in which I work,
I thought I should tell you a

229
00:18:32 --> 00:18:41
bit about it.
OK, so now back to reviewing

230
00:18:41 --> 00:18:47
for the exam.
So, I'm going to basically list

231
00:18:47 --> 00:18:49
topics.
And, if time permits,

232
00:18:49 --> 00:18:53
I will say a few things about
problems from practice exam 3B.

233
00:18:53 --> 00:18:56
I'm hoping that you have it or
your neighbor has it,

234
00:18:56 --> 00:18:59
or you can somehow get it.
Anyway, given time,

235
00:18:59 --> 00:19:04
I'm not sure how much I will
say about the problems in and of

236
00:19:04 --> 00:19:08
themselves.
OK, so the main thing to know

237
00:19:08 --> 00:19:13
about this exam is how to set up
and evaluate double integrals

238
00:19:13 --> 00:19:17
and line integrals.
OK, if you know how to do these

239
00:19:17 --> 00:19:20
two things, then you are in much
better shape than if you don't.

240
00:19:20 --> 00:19:26
 
 

241
00:19:26 --> 00:19:43
And -- So, the first thing
we've seen, just to write it

242
00:19:43 --> 00:19:55
down, there's two main objects.
And, it's kind of important to

243
00:19:55 --> 00:19:57
not confuse them with each
other.

244
00:19:57 --> 00:20:02
OK, there's double integrals of
our regions of some quantity,

245
00:20:02 --> 00:20:06
dA, 
and the other one is the line

246
00:20:06 --> 00:20:11
integral along a curve of a
vector field,

247
00:20:11 --> 00:20:17
F.dr or F.Mds depending on
whether it's work or flux that

248
00:20:17 --> 00:20:21
we are trying to do.
And, so we should know how to

249
00:20:21 --> 00:20:24
set up these things and how to
evaluate them.

250
00:20:24 --> 00:20:27
And, roughly speaking,
in this one you start by

251
00:20:27 --> 00:20:32
drawing a picture of the region,
then deciding which way you

252
00:20:32 --> 00:20:34
will integrate it.
It could be dx dy, 

253
00:20:34 --> 00:20:37
dy dx, 
r dr d theta, 

254
00:20:37 --> 00:20:41
and then you will set up the
bound carefully by slicing it

255
00:20:41 --> 00:20:45
and studying how the bounds for
the inner variable depend on the

256
00:20:45 --> 00:20:51
outer variable.
So, the first topic will be

257
00:20:51 --> 00:20:57
setting up double integrals.
And so, remember,

258
00:20:57 --> 00:21:03
OK, so maybe I should make this
more explicit.

259
00:21:03 --> 00:21:12
We want to draw a picture of R
and take slices in the chosen

260
00:21:12 --> 00:21:18
way so that we get an iterated
integral.

261
00:21:18 --> 00:21:25
OK, so let's do just a quick
example.

262
00:21:25 --> 00:21:38
So, if I look at problem one on
the exam 3B,

263
00:21:38 --> 00:21:43
it says to look at the line
integral from zero to one,

264
00:21:43 --> 00:21:46
line integral from x to 2x of
possibly something,

265
00:21:46 --> 00:21:50
but dy dx.
And it says,

266
00:21:50 --> 00:21:58
let's look at how we would set
this up the other way around by

267
00:21:58 --> 00:22:03
exchanging x and y.
So, we should get to something

268
00:22:03 --> 00:22:06
that will be the same integral
dx dy.

269
00:22:06 --> 00:22:09
I mean, if you have a function
of x and y, then it will be the

270
00:22:09 --> 00:22:11
same function.
But, of course,

271
00:22:11 --> 00:22:14
the bounds change.
So, how do we exchange the

272
00:22:14 --> 00:22:17
order of integration?
Well, the only way to do it

273
00:22:17 --> 00:22:20
consistently is to draw a
picture.

274
00:22:20 --> 00:22:23
So, let's see,
what does this mean?

275
00:22:23 --> 00:22:28
Here, it means we integrate
from y equals x to y equals 2x,

276
00:22:28 --> 00:22:32
x between zero and one.
So, we should draw a picture.

277
00:22:32 --> 00:22:35
The lower bound for y is y
equals x.

278
00:22:35 --> 00:22:41
So, let's draw y equals x.
That seems to be here.

279
00:22:41 --> 00:22:47
And, we'll go up to y equals
2x, which is a line also but

280
00:22:47 --> 00:22:52
with bigger slope.
And then, all right,

281
00:22:52 --> 00:22:58
so for each value of x,
my origin will go from x to 2x.

282
00:22:58 --> 00:23:03
Well, and I do this for all
values of x that go to x equals

283
00:23:03 --> 00:23:06
one.
So, I stop at x equals one,

284
00:23:06 --> 00:23:10
which is here.
And then, my region is

285
00:23:10 --> 00:23:15
something like this.
OK, so this point here,

286
00:23:15 --> 00:23:21
in case you are wondering,
well, when x equals one,

287
00:23:21 --> 00:23:27
y is one.
And that point here is one, two.

288
00:23:27 --> 00:23:29
OK, any questions about that so
far?

289
00:23:29 --> 00:23:33
OK, so somehow that's the first
kill, when you see an integral,

290
00:23:33 --> 00:23:36
how to figure out what it
means, how to draw the region.

291
00:23:36 --> 00:23:39
And then there's a converse
scale which is given the region,

292
00:23:39 --> 00:23:42
how to set up the integral for
it.

293
00:23:42 --> 00:23:46
So, if we want to set up
instead dx dy,

294
00:23:46 --> 00:23:50
then it means we are going to
actually look at the converse

295
00:23:50 --> 00:23:54
question which is,
for a given value of y, 

296
00:23:54 --> 00:23:57
what is the range of values of
x?

297
00:23:57 --> 00:24:01
OK, so if we fix y,
well, where do we enter the

298
00:24:01 --> 00:24:04
region, and where do we leave
it?

299
00:24:04 --> 00:24:08
So, we seem to enter on this
side, and we seem to leave on

300
00:24:08 --> 00:24:10
that side.
At least that seems to be true

301
00:24:10 --> 00:24:12
for the first few values of y
that I choose.

302
00:24:12 --> 00:24:16
But, hey, if I take a larger
value of y, then I will enter on

303
00:24:16 --> 00:24:19
the side, and I will leave on
this vertical side,

304
00:24:19 --> 00:24:22
not on that one.
So, I seem to have two

305
00:24:22 --> 00:24:28
different things going on.
OK, the place where enter my

306
00:24:28 --> 00:24:38
region is always y equals 2x,
which is the same as x equals y

307
00:24:38 --> 00:24:45
over two.
So, x seems to always start at

308
00:24:45 --> 00:24:51
y over two.
But, where I leave to be either

309
00:24:51 --> 00:24:55
x equals y, or here,
x equals y.

310
00:24:55 --> 00:24:57
And, that depends on the value
of y.

311
00:24:57 --> 00:24:59
So, in fact,
I have to break this into two

312
00:24:59 --> 00:25:03
different integrals.
I have to treat separately the

313
00:25:03 --> 00:25:07
case where y is between zero and
one, and between one and two.

314
00:25:07 --> 00:25:15
So, what I do in that case is I
just make two integrals.

315
00:25:15 --> 00:25:18
So, I say, both of them start
at y over two.

316
00:25:18 --> 00:25:22
But, in the first case,
we'll stop at x equals y.

317
00:25:22 --> 00:25:30
In the second case,
we'll stop at x equals one.

318
00:25:30 --> 00:25:31
OK, and now,
what are the values of y for

319
00:25:31 --> 00:25:34
each case?
Well, the first case is when y

320
00:25:34 --> 00:25:38
is between zero and one.
The second case is when y is

321
00:25:38 --> 00:25:40
between one and two,
which I guess this picture now

322
00:25:40 --> 00:25:44
is completely unreadable,
but hopefully you've been

323
00:25:44 --> 00:25:48
following what's going on,
or else you can see it in the

324
00:25:48 --> 00:25:53
solutions to the problem.
And, so that's our final answer.

325
00:25:53 --> 00:26:01
OK, any questions about how to
set up double integrals in xy

326
00:26:01 --> 00:26:04
coordinates?
No?

327
00:26:04 --> 00:26:07
OK, who feels comfortable with
this kind of problem?

328
00:26:07 --> 00:26:11
OK, good.
I'm happy to see the vast

329
00:26:11 --> 00:26:16
majority.
So, the bad news is we have to

330
00:26:16 --> 00:26:23
be able to do it not only in xy
coordinates, but also in polar

331
00:26:23 --> 00:26:27
coordinates.
So, when you go to polar

332
00:26:27 --> 00:26:32
coordinates, basically all you
have to remember on the side of

333
00:26:32 --> 00:26:36
integrand is that x becomes r
cosine theta.

334
00:26:36 --> 00:26:45
Y becomes r sine theta.
And, dx dy becomes r dr d theta.

335
00:26:45 --> 00:26:49
In terms of how you slice for
your region, well,

336
00:26:49 --> 00:26:52
you will be integrating first
over r.

337
00:26:52 --> 00:26:57
So, that means what you're
doing is you're fixing the value

338
00:26:57 --> 00:26:59
of theta.
And, for that value of theta,

339
00:26:59 --> 00:27:03
you ask yourself,
for what range of values of r

340
00:27:03 --> 00:27:06
am I going to be inside my
origin?

341
00:27:06 --> 00:27:09
So, if my origin looks like
this, then for this value of

342
00:27:09 --> 00:27:13
theta, r would go from zero to
whatever this distance is.

343
00:27:13 --> 00:27:16
And of course I have to find
how this distance depends on

344
00:27:16 --> 00:27:18
theta.
And then, I will find the

345
00:27:18 --> 00:27:20
extreme values of theta.
Now, of course,

346
00:27:20 --> 00:27:22
is the origin is really looking
like this, then you're not going

347
00:27:22 --> 00:27:25
to do it in polar coordinates.
But, if it's like a circle or a

348
00:27:25 --> 00:27:27
half circle, or things like
that,

349
00:27:27 --> 00:27:31
then even if a problem doesn't
tell you to do it in polar

350
00:27:31 --> 00:27:34
coordinates you might want to
seriously consider it.

351
00:27:34 --> 00:27:38
OK, so I'm not going to do it
but problem two in the practice

352
00:27:38 --> 00:27:43
exam is a good example of doing
something in polar coordinates.

353
00:27:43 --> 00:27:50
OK, 
so in terms of things that we

354
00:27:50 --> 00:27:56
do with double integrals,
there's a few formulas that I'd

355
00:27:56 --> 00:28:00
like you to remember about
applications that we've seen of

356
00:28:00 --> 00:28:04
double integrals.
So, quantities that we can

357
00:28:04 --> 00:28:10
compute with double integrals
include things like the area of

358
00:28:10 --> 00:28:13
region,
its mass if it has a density, 

359
00:28:13 --> 00:28:16
the average value of some
function,

360
00:28:16 --> 00:28:19
for example, 
the average value of the x and

361
00:28:19 --> 00:28:22
y coordinates,
which we called the center of

362
00:28:22 --> 00:28:31
mass or moments of inertia.
So, these are just formulas to

363
00:28:31 --> 00:28:35
remember.
So, for example,

364
00:28:35 --> 00:28:40
the area of region is the
double integral of just dA,

365
00:28:40 --> 00:28:44
or if it helps you,
one dA if you want.

366
00:28:44 --> 00:28:47
You are integrating the
function 1.

367
00:28:47 --> 00:28:49
You have to remember formulas
for mass,

368
00:28:49 --> 00:28:54
for the average value of a
function is the F bar,

369
00:28:54 --> 00:29:05
in particular x bar y bar, 
which is the center of mass, 

370
00:29:05 --> 00:29:14
and the moment of inertia.
OK, so the polar moment of

371
00:29:14 --> 00:29:18
inertia, which is moment of
inertia about the origin.

372
00:29:18 --> 00:29:22
OK, so that's double integral
of x squared plus y squared,

373
00:29:22 --> 00:29:27
density dA, 
but also moments of inertia

374
00:29:27 --> 00:29:33
about the x and y axis,
which are given by just taking

375
00:29:33 --> 00:29:36
one of these guys.
Don't worry about moments of

376
00:29:36 --> 00:29:39
inertia about an arbitrary line.
I will ask you for a moment of

377
00:29:39 --> 00:29:42
inertia for some weird line or
something like that.

378
00:29:42 --> 00:29:47
OK, but these you should know.
Now, what if you somehow,

379
00:29:47 --> 00:29:49
on the spur of the moment,
you forget, what's the formula

380
00:29:49 --> 00:29:51
for moment of inertia?
Well, I mean,

381
00:29:51 --> 00:29:54
I prefer if you know,
but if you have a complete

382
00:29:54 --> 00:29:56
blank in your memory,
there will still be partial

383
00:29:56 --> 00:29:59
credit were setting up the
bounds and everything else.

384
00:29:59 --> 00:30:01
So, 
the general rule for the exam

385
00:30:01 --> 00:30:04
will be if you're stuck in a
calculation or you're missing a

386
00:30:04 --> 00:30:08
little piece of the puzzle,
try to do as much as you can.

387
00:30:08 --> 00:30:10
In particular,
try to at least set up the

388
00:30:10 --> 00:30:15
bounds of the integral.
There will be partial credit

389
00:30:15 --> 00:30:21
for that always.
So, while we're at it about

390
00:30:21 --> 00:30:26
grand rules, how about
evaluation?

391
00:30:26 --> 00:30:31
How about evaluating integrals?
So, once you've set it up,

392
00:30:31 --> 00:30:33
you have to sometimes compute
it.

393
00:30:33 --> 00:30:36
First of all,
check just in case the problem

394
00:30:36 --> 00:30:40
says set up but do not evaluate.
Then, don't waste your time

395
00:30:40 --> 00:30:45
evaluating it.
If a problem says to compute

396
00:30:45 --> 00:30:50
it, then you have to compute it.
So, what kinds of integration

397
00:30:50 --> 00:30:54
techniques do you need to know?
So, you need to know, 

398
00:30:54 --> 00:30:57
you must know, 
well, how to integrate the

399
00:30:57 --> 00:31:01
usual functions like one over x
or x to the n,

400
00:31:01 --> 00:31:05
or exponential, 
sine, cosine, 

401
00:31:05 --> 00:31:08
things like that, 
OK, so the usual integrals.

402
00:31:08 --> 00:31:16
You must know what I will call
easy trigonometry.

403
00:31:16 --> 00:31:17
OK, I don't want to give you a
complete list.

404
00:31:17 --> 00:31:20
And the more you ask me about
which ones are on the list,

405
00:31:20 --> 00:31:22
the more I will add to the
list.

406
00:31:22 --> 00:31:26
But, those that you know that
you should know,

407
00:31:26 --> 00:31:28
you should know.
Those that you think you

408
00:31:28 --> 00:31:31
shouldn't know,
you don't have to know because

409
00:31:31 --> 00:31:36
I will say what I will say soon.
You should know also

410
00:31:36 --> 00:31:41
substitution,
how to set U equals something,

411
00:31:41 --> 00:31:45
and then see,
oh, this becomes u times du,

412
00:31:45 --> 00:31:50
and so substitution method.
What do I mean by easy

413
00:31:50 --> 00:31:52
trigonometrics?
Well, certainly you should know

414
00:31:52 --> 00:31:54
how to ingrate sine.
You should know how to

415
00:31:54 --> 00:31:57
integrate cosine.
You should be aware that sine

416
00:31:57 --> 00:32:01
squared plus cosine squared
simplifies to one.

417
00:32:01 --> 00:32:03
And, you should be aware of
general things like that.

418
00:32:03 --> 00:32:06
I would like you to know,
maybe, the double angles,

419
00:32:06 --> 00:32:09
sine 2x and cosine 2x.
Know what these are, 

420
00:32:09 --> 00:32:12
and the kinds of the easy
things you can do with that,

421
00:32:12 --> 00:32:16
also things that involve
substitution setting like U

422
00:32:16 --> 00:32:19
equals sine T or U equals cosine
T.

423
00:32:19 --> 00:32:21
I mean, let me,
instead, give an example of

424
00:32:21 --> 00:32:25
hard trig that you don't need to
know, and then I will answer.

425
00:32:25 --> 00:32:34
OK, so, not needed on Thursday;
it doesn't mean that I don't

426
00:32:34 --> 00:32:37
want you to know them.
I would love you to know every

427
00:32:37 --> 00:32:41
single integral formula.
But, that shouldn't be your top

428
00:32:41 --> 00:32:44
priority.
So, you don't need to know

429
00:32:44 --> 00:32:47
things like hard trigonometric
ones.

430
00:32:47 --> 00:32:52
So, let me give you an example.
OK, so if I ask you to do this

431
00:32:52 --> 00:32:55
one, then actually I will give
you maybe, you know,

432
00:32:55 --> 00:32:59
I will reprint the formula from
the notes or something like

433
00:32:59 --> 00:33:02
that.
OK, so that one you don't need

434
00:33:02 --> 00:33:04
to know.
I would love if you happen to

435
00:33:04 --> 00:33:07
know it, but if you need it,
it will be given to you.

436
00:33:07 --> 00:33:13
So, these kinds of things that
you cannot compute by any easy

437
00:33:13 --> 00:33:16
method.
And, integration by parts,

438
00:33:16 --> 00:33:21
I believe that I successfully
test-solved all the problems

439
00:33:21 --> 00:33:26
without doing any single
integration by parts.

440
00:33:26 --> 00:33:29
Again, in general,
it's something that I would

441
00:33:29 --> 00:33:33
like you to know,
but it shouldn't be a top

442
00:33:33 --> 00:33:40
priority for this week.
OK, sorry, you had a question,

443
00:33:40 --> 00:33:42
or?
Inverse trigonometric

444
00:33:42 --> 00:33:45
functions: let's say the most
easy ones.

445
00:33:45 --> 00:33:50
I would like you to know the
easiest inverse trig functions,

446
00:33:50 --> 00:33:56
but not much.
OK, OK, so be aware that these

447
00:33:56 --> 00:34:04
functions exist,
but it's not a top priority.

448
00:34:04 --> 00:34:06
I should say,
the more I tell you I don't

449
00:34:06 --> 00:34:08
need you to know,
the more your physics and other

450
00:34:08 --> 00:34:11
teachers might complain that,
oh, these guys don't know how

451
00:34:11 --> 00:34:12
to integrate.
So, try not to forget

452
00:34:12 --> 00:34:19
everything.
But, yes?

453
00:34:19 --> 00:34:22
No, no, here I just mean for
evaluating just a single

454
00:34:22 --> 00:34:24
variable integral.
I will get to change variables

455
00:34:24 --> 00:34:27
and Jacobian soon,
but I'm thinking of this as a

456
00:34:27 --> 00:34:29
different topic.
What I mean by this one is if

457
00:34:29 --> 00:34:32
I'm asking you to integrate,
I don't know,

458
00:34:32 --> 00:34:37
what's a good example?
Zero to one t dt over square

459
00:34:37 --> 00:34:42
root of one plus t squared,
then you should think of maybe

460
00:34:42 --> 00:34:44
substituting u equals one plus t
squared,

461
00:34:44 --> 00:34:55
and then it becomes easier.
OK, so this kind of trig,

462
00:34:55 --> 00:35:00
that's what I have in mind here
specifically.

463
00:35:00 --> 00:35:02
And again, 
if you're stuck, 

464
00:35:02 --> 00:35:05
in particular,
if you hit this dreaded guy,

465
00:35:05 --> 00:35:09
and you don't actually have a
formula giving you what it is,

466
00:35:09 --> 00:35:12
it means one of two things.
One is something's wrong with

467
00:35:12 --> 00:35:13
your solution.
The other option is something

468
00:35:13 --> 00:35:16
is wrong with my problem.
So, either way,

469
00:35:16 --> 00:35:22
check quickly what you've done
it if you can't find a mistake,

470
00:35:22 --> 00:35:27
then just move ahead to the
next problem.

471
00:35:27 --> 00:35:30
Which one, this one?
Yeah, 

472
00:35:30 --> 00:35:32
I mean if you can do it, 
if you know how to do it, 

473
00:35:32 --> 00:35:33
which everything is fair:
I mean,

474
00:35:33 --> 00:35:36
generally speaking, 
give enough of it so that you

475
00:35:36 --> 00:35:38
found the solution by yourself,
not like, 

476
00:35:38 --> 00:35:43
you know, it didn't somehow
come to you by magic.

477
00:35:43 --> 00:35:47
But, yeah, if you know how to
integrate this without doing the

478
00:35:47 --> 00:35:49
substitution,
that's absolutely fine by me.

479
00:35:49 --> 00:35:53
Just show enough work.
The general rule is show enough

480
00:35:53 --> 00:35:58
work that we see that you knew
what you are doing.

481
00:35:58 --> 00:36:02
OK, now another thing we've
seen with double integrals is

482
00:36:02 --> 00:36:05
how to do more complicated
changes of variables.

483
00:36:05 --> 00:36:18
 
 

484
00:36:18 --> 00:36:23
So, when you want to replace x
and y by some variables,

485
00:36:23 --> 00:36:28
u and v, given by some formulas
in terms of x and y.

486
00:36:28 --> 00:36:33
So, you need to remember
basically how to do them.

487
00:36:33 --> 00:36:36
So, you need to remember that
the method consists of three

488
00:36:36 --> 00:36:43
steps.
So, one is you have to find the

489
00:36:43 --> 00:36:46
Jacobian.
And, you can choose to do

490
00:36:46 --> 00:36:50
either this Jacobian or the
inverse one depending on what's

491
00:36:50 --> 00:36:53
easiest given what you're given.
You don't have to worry about

492
00:36:53 --> 00:36:55
solving for things the other way
around.

493
00:36:55 --> 00:36:58
Just compute one of these
Jacobians.

494
00:36:58 --> 00:37:06
And then, the rule is that du
dv is absolute value of the

495
00:37:06 --> 00:37:12
Jacobian dx dy.
So, that takes care of dx dy,

496
00:37:12 --> 00:37:18
how to convert that into du dv.
The second thing to know is

497
00:37:18 --> 00:37:20
that,
well, 

498
00:37:20 --> 00:37:25
you need to of course
substitute any x and y's in the

499
00:37:25 --> 00:37:32
integrand to convert them to u's
and v's so that you have a valid

500
00:37:32 --> 00:37:36
integrand involving only u and
v.

501
00:37:36 --> 00:37:51
And then, the last part is
setting up the bounds.

502
00:37:51 --> 00:37:54
And you see that,
probably you seen on P-sets and

503
00:37:54 --> 00:37:58
an example we did in the lecture
that this can be complicated.

504
00:37:58 --> 00:38:00
But now, in real life,
you do this actually to

505
00:38:00 --> 00:38:02
simplify the integrals.
So, 

506
00:38:02 --> 00:38:04
probably the one that will be
there on Thursday,

507
00:38:04 --> 00:38:07
if there's a problem about that
on Thursday,

508
00:38:07 --> 00:38:10
it will be a situation where
the bounds that you get after

509
00:38:10 --> 00:38:13
changing variables are
reasonably easy.

510
00:38:13 --> 00:38:15
OK, I'm not saying that it will
be completely obvious

511
00:38:15 --> 00:38:17
necessarily, but it will be a
fairly easy situation.

512
00:38:17 --> 00:38:22
So, the general method is you
look at your region,

513
00:38:22 --> 00:38:25
R, and it might have various
sides.

514
00:38:25 --> 00:38:29
Well, on each side you ask
yourself, what do I know about x

515
00:38:29 --> 00:38:33
and y, and how to convert that
in terms of u and v?

516
00:38:33 --> 00:38:37
And maybe you'll find that the
equation might be just u equals

517
00:38:37 --> 00:38:39
zero for example,
or u equals v,

518
00:38:39 --> 00:38:42
or something like that.
And then, it's up to you to

519
00:38:42 --> 00:38:46
decide what you want to do.
But, maybe the easiest usually

520
00:38:46 --> 00:38:49
is to draw a new picture in
terms of u and v coordinates of

521
00:38:49 --> 00:38:53
what your region will look like
in the new coordinates.

522
00:38:53 --> 00:38:55
It might be that it will
actually much easier.

523
00:38:55 --> 00:39:00
It should be easier looking
than what you started with.

524
00:39:00 --> 00:39:05
OK, so that's the general idea.
There is one change of variable

525
00:39:05 --> 00:39:09
problem on each of the two
practice exams to give you a

526
00:39:09 --> 00:39:13
feeling for what's realistic.
The problem that's on practice

527
00:39:13 --> 00:39:18
exam 3B actually is on the hard
side of things because the

528
00:39:18 --> 00:39:21
question is kind of hidden in a
way.

529
00:39:21 --> 00:39:25
So, if you look at problem six,
you might find that it's not

530
00:39:25 --> 00:39:28
telling you very clearly what
you have to do.

531
00:39:28 --> 00:39:34
That's because it was meant to
be the hardest problem on that

532
00:39:34 --> 00:39:37
test.
But, once you've reduced it to

533
00:39:37 --> 00:39:41
an actual change of variables
problem, I expect you to be able

534
00:39:41 --> 00:39:44
to know how to do it.
And, on practice exam 3A,

535
00:39:44 --> 00:39:48
there's also,
I think it's problem five on

536
00:39:48 --> 00:39:52
the other practice exam.
And, that one is actually

537
00:39:52 --> 00:39:55
pretty standard and
straightforward.

538
00:39:55 --> 00:40:00
OK, time to move on, sorry.
So, we've also seen about line

539
00:40:00 --> 00:40:00
integrals.

540
00:40:00 --> 00:40:21

541
00:40:21 --> 00:40:30
OK, 
so line integrals, 

542
00:40:30 --> 00:40:33
so the main thing to know about
them,

543
00:40:33 --> 00:40:37
so the line integral for work, 
which is line integral of F.dr, 

544
00:40:37 --> 00:40:40
so let's say that your vector
field has components,

545
00:40:40 --> 00:40:49
M and N.
So, the line integral for work

546
00:40:49 --> 00:40:57
becomes in coordinates integral
of Mdx plus Ndy while we've also

547
00:40:57 --> 00:41:05
seen line integral for flux.
So, line integral of F.n ds

548
00:41:05 --> 00:41:13
becomes the integral along C
just to make sure that I give it

549
00:41:13 --> 00:41:18
to you correctly.
So, remember that just,

550
00:41:18 --> 00:41:22
I don't want to make the
mistake in front of you.

551
00:41:22 --> 00:41:30
So, T ds is dx, dy.
And, the normal vector,

552
00:41:30 --> 00:41:36
so, T ds goes along the curve.
Nds goes clockwise

553
00:41:36 --> 00:41:41
perpendicular to the curve.
So, it's going to be,

554
00:41:41 --> 00:41:48
well, it's going to be dy and
negative dx.

555
00:41:48 --> 00:42:00
So, you will be integrating
negative Ndx plus Mdy.

556
00:42:00 --> 00:42:04
OK, see, if you are blanking
and don't remember the signs,

557
00:42:04 --> 00:42:07
then you can just draw this
picture and make sure that you

558
00:42:07 --> 00:42:10
get it right.
So, you should know a little

559
00:42:10 --> 00:42:14
bit about geometric
interpretation and how to see

560
00:42:14 --> 00:42:17
easily that it's going to be
zero in some cases.

561
00:42:17 --> 00:42:21
But, mostly you should know how
to compute, set up and compute

562
00:42:21 --> 00:42:23
these things.
So, what do we do when we are

563
00:42:23 --> 00:42:24
here?
Well, it's year,

564
00:42:24 --> 00:42:27
we have both x and y together,
but we want to,

565
00:42:27 --> 00:42:30
because it's the line integral,
there should be only one

566
00:42:30 --> 00:42:34
variable.
So, the important thing to know

567
00:42:34 --> 00:42:39
is we want to reduce everything
to a single parameter.

568
00:42:39 --> 00:42:55
OK, so the evaluation method is
always by reducing to a single

569
00:42:55 --> 00:43:01
parameter.
So, for example, 

570
00:43:01 --> 00:43:06
maybe x and y are both
functions of some variable,

571
00:43:06 --> 00:43:10
t,
and then express everything in

572
00:43:10 --> 00:43:18
terms of some integral of,
some quantity involving t dt.

573
00:43:18 --> 00:43:21
It could be that you will just
express everything in terms of x

574
00:43:21 --> 00:43:24
or in terms of y,
or in terms of some angle or

575
00:43:24 --> 00:43:26
something.
It's up to you to choose how to

576
00:43:26 --> 00:43:29
parameterize things.
And then, when you're there,

577
00:43:29 --> 00:43:33
it's a usual one variable
integral with a single variable

578
00:43:33 --> 00:43:36
in there.
OK, so that's the general

579
00:43:36 --> 00:43:40
method of calculation,
but we've seen a shortcut for

580
00:43:40 --> 00:43:45
work when we can show that the
field is the gradient of

581
00:43:45 --> 00:43:48
potential.
So, 

582
00:43:48 --> 00:43:55
one thing to know is if the
curl of F,

583
00:43:55 --> 00:44:01
which is an x minus My happens
to be zero,

584
00:44:01 --> 00:44:03
well, 
and now I can say, 

585
00:44:03 --> 00:44:06
and the domain is simply
connected,

586
00:44:06 --> 00:44:11
or if the field is defined
everywhere,

587
00:44:11 --> 00:44:19
then F is actually a gradient
field.

588
00:44:19 --> 00:44:22
So, that means, 
just to make it more concrete, 

589
00:44:22 --> 00:44:26
that means we can find a
function little f called the

590
00:44:26 --> 00:44:30
potential such that its
derivative respect to x is M,

591
00:44:30 --> 00:44:32
and its derivative with respect
to Y is N.

592
00:44:32 --> 00:44:37
We can solve these two
conditions for the same

593
00:44:37 --> 00:44:42
function, f, simultaneously.
And, how do we find this

594
00:44:42 --> 00:44:46
function, little f?
OK, so that's the same as

595
00:44:46 --> 00:44:50
saying that the field,
big F, is the gradient of

596
00:44:50 --> 00:44:52
little f.
And, how do we find this

597
00:44:52 --> 00:44:54
function, little f?
Well, we've seen two methods.

598
00:44:54 --> 00:44:58
One of them involves computing
a line integral from the origin

599
00:44:58 --> 00:45:02
to a point in the plane by going
first along the x axis,

600
00:45:02 --> 00:45:05
then vertically.
The other method was to first

601
00:45:05 --> 00:45:09
figure out what this one tells
us by integrating it with

602
00:45:09 --> 00:45:12
respect to x.
And then, we differentiate our

603
00:45:12 --> 00:45:17
answer with respect to y,
and we compare with that to get

604
00:45:17 --> 00:45:20
the complete answer.
OK, so I is that relevant?

605
00:45:20 --> 00:45:22
Well, 
first of all it's relevant in

606
00:45:22 --> 00:45:25
physics,
but it's also relevant just to

607
00:45:25 --> 00:45:29
calculation of line integrals
because we see the fundamental

608
00:45:29 --> 00:45:34
theorem of calculus for line
integrals which says if we are

609
00:45:34 --> 00:45:39
integrating a gradient field and
we know what the potential is.

610
00:45:39 --> 00:45:43
Then, we just have to,
well, the line integral is just

611
00:45:43 --> 00:45:46
the change in value of a
potential.

612
00:45:46 --> 00:45:49
OK, so we take the value of a
potential at the starting point,

613
00:45:49 --> 00:45:52
sorry, we take value potential
at the endpoint minus the value

614
00:45:52 --> 00:45:58
at the starting point.
And, that will give us the line

615
00:45:58 --> 00:46:00
integral, OK?
So, important:

616
00:46:00 --> 00:46:05
this is only for work.
There's no statement like that

617
00:46:05 --> 00:46:09
for flux, OK,
so don't tried to fly this in a

618
00:46:09 --> 00:46:11
problem about flux.
I mean, usually, 

619
00:46:11 --> 00:46:13
if you look at the practice
exams,

620
00:46:13 --> 00:46:17
you will see it's pretty clear
that there's one problem in

621
00:46:17 --> 00:46:20
which you are supposed to do
things this way.

622
00:46:20 --> 00:46:25
It's kind of a dead giveaway,
but it's probably not too bad.

623
00:46:25 --> 00:46:29
OK, and the other thing we've
seen, so I mentioned it at the

624
00:46:29 --> 00:46:32
beginning but let me mention it
again.

625
00:46:32 --> 00:46:36
To compute things, 
Green's theorem, 

626
00:46:36 --> 00:46:42
let's just compute, 
well, let us forget, 

627
00:46:42 --> 00:46:45
sorry, find the value of a line
integral along the closed curve

628
00:46:45 --> 00:46:47
by reducing it to double
integral.

629
00:46:47 --> 00:46:55
So, 
the one for work says -- --

630
00:46:55 --> 00:46:59
this,
and you should remember that in

631
00:46:59 --> 00:47:01
there,
so C is a closed curve that

632
00:47:01 --> 00:47:05
goes counterclockwise,
and R is the region inside.

633
00:47:05 --> 00:47:08
So, the way you would,
if you had to compute both

634
00:47:08 --> 00:47:10
sides separately,
you would do them in extremely

635
00:47:10 --> 00:47:12
different ways,
right?

636
00:47:12 --> 00:47:15
This one is a line integral.
So, you use the method to

637
00:47:15 --> 00:47:18
explain here,
namely, you express x and y in

638
00:47:18 --> 00:47:22
terms of a single variable.
See that you're doing a circle.

639
00:47:22 --> 00:47:24
I want to see a theta.
I don't want to see an R.

640
00:47:24 --> 00:47:27
R is not a variable.
You are on the circle.

641
00:47:27 --> 00:47:30
This one is a double integral.
So, if you are doing it,

642
00:47:30 --> 00:47:32
say, on a disk,
you would have both R and theta

643
00:47:32 --> 00:47:34
if you're using polar
coordinates.

644
00:47:34 --> 00:47:37
You would have both x and y.
Here, you have two variables of

645
00:47:37 --> 00:47:40
integration.
Here, you should have only one

646
00:47:40 --> 00:47:42
after you parameterize the
curve.

647
00:47:42 --> 00:47:46
And, the fact that it stays
curl F, I mean,

648
00:47:46 --> 00:47:51
curl F is just Nx-My is just
like any function of x and y.

649
00:47:51 --> 00:47:54
OK, the fact that we called it
curl F doesn't change how you

650
00:47:54 --> 00:47:56
compute it.
You have first to compute the

651
00:47:56 --> 00:47:58
curl of F.
Say you find,

652
00:47:58 --> 00:48:00
I don't know,
xy minus x squared,

653
00:48:00 --> 00:48:04
well, it becomes just the usual
double integral of the usual

654
00:48:04 --> 00:48:09
function xy minus x squared.
There's nothing special to it

655
00:48:09 --> 00:48:15
because it's a curl.
And, the other one is the

656
00:48:15 --> 00:48:21
counterpart for flux.
So, it says this,

657
00:48:21 --> 00:48:25
and remember this is mx plus
ny.

658
00:48:25 --> 00:48:27
I mean, what's important about
these statements is not only

659
00:48:27 --> 00:48:30
remembering, you know,
if you just know this formula

660
00:48:30 --> 00:48:32
by heart,
you are still in trouble

661
00:48:32 --> 00:48:35
because you need to know what
actually the symbols in here

662
00:48:35 --> 00:48:37
mean.
So, you should remember,

663
00:48:37 --> 00:48:40
what is this line integral,
and what's the divergence of a

664
00:48:40 --> 00:48:47
field?
So, just something to remember.

665
00:48:47 --> 00:48:51
And, so I guess I'll let you
figure out practice problems

666
00:48:51 --> 00:48:54
because it's time,
but I think that's basically

667
00:48:54 --> 00:48:59
the list of all we've seen.
And, well, that should be it.

668
00:48:59 --> 00:49:60